If we want to integrate over a region which doesn't split nicely along lines parallel to the coordinate axes, we can split the region up along other lines or curves.

For example, consider integrating over the region bounded by the hyperbolas and and the lines and .This region naturally splits along hyperbolas of the form where ranges from 1 to 3 and lines of the form where ranges from to 2. One convenient way of describing this family of curves and lines we're using to slice up the region is to write the region as the image of the rectangle under the inverse of the transformation .

To integrate over the region in the -plane using this subdivision, we need to find the area of each small piece. Since the areas of the rectangles in the -plane are easier to deal with, we find the areas of the piece in the -plane by the area of the corresponding rectangle by the area distortion factor of the transformation .This local area distortion factor, or Jacobian determinant is the absolute value of the determinant of the Jacobian matrix .This is because the area distortion factor of a linear transformation is the absolute value of its determinant, and a differentiable function is essentially linear when zoomed far enough in. Thus we arrive at the formula

where \left|\frac{\partial \mathbf{T}(u,v)}{\partial (u,v)} \right| is notation for the absolute value of the determinant of the transformation \frac{\partial \mathbf{T}(u,v)}{\partial (u,v)}.

Since the area distortion factor of \mathbf{T} is the of the area distortion factor of \mathbf{T}^{-1}, we can work out that \left|\frac{\partial \mathbf{T}(u,v)}{\partial (u,v)} \right| = \left|\begin{array}{cc} x & y \\ 1/x & -y/x^2 \end{array} \right|^{-1} = (2y/x)^{-1} = \frac{1}{2v}, which implies

Let's apply the Jacobian transformation idea to a familiar problem: the area enclosed by a circle.

Exercise Use the map (u,v) \mapsto (u \cos v, u \sin v) from the rectangle [0,1] \times [0,2\pi] to the unit disk, and calculate the Jacobian for this transformation. Use your result to integrate 1 over the unit disk and confirm that the result is equal to the area of the unit disk.

Solution.The Jacobian of the given transformation is

\begin{align*}\left|\begin{array}{cc} \cos v & \sin v \\ -u\sin v & u \cos v \end{array} \right| = u(\cos^2 v + \sin^2 v) = u.\end{align*}

So the area of the disk is equal to

\begin{align*}\int_0^1 \int_0^{2\pi} u \, \mathrm{d}u \mathrm{d}v = \boxed{\pi}.\end{align*}

Congratulations! You have finished the Data Gymnasia multivariable calculus course.